Timeline for Reference request: Optimal $L^p$-decay for nonhomogenous heat equation in $\mathbb R^d$
Current License: CC BY-SA 3.0
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| Apr 24, 2015 at 10:27 | comment | added | Juhana Siljander | Well, this is more or less, what I wanted to ask. The method you propose is essentially the one I originally studied. We get a decay result, but with a condition on $p$, which seems unnatural for me. For instance, I would think that $u$ has decay even in $L^\infty$ norm, provided $f$ has an appropriate decay behavior. So my question is whether the result you obtain - with the assumption $\frac1q-\frac1p < \frac2d$ - is optimal/sharp in terms of the range of $p$? Or is it just that with this method we cannot get any better, but there is another argument which improves the result? | |
| Apr 22, 2015 at 7:20 | comment | added | ifw | @Juhana It is unlikely that similar estimates hold if $p$ is too large. This is because the estimates under the integral (in the first displayed line after the first grey block) are pointwise sharp, and integration cannot improve the situation (e.g., due to cancellation) as the heat semigroup is positive preserving. | |
| Apr 21, 2015 at 9:40 | comment | added | Juhana Siljander | What I meant is that the condition $\frac1q-\frac1p < \frac2d$ seems to give an upper bound for $p$ in terms of $q$ and $d$ (at least if $q$ is small enough). So what happens if $p$ is larger than this number? | |
| Apr 20, 2015 at 6:29 | comment | added | ifw | @Juhana For $\gamma>1$, there is a decay of $\displaystyle (1+t)^{-\frac{d}2\left(\frac1q-\frac1p\right)}$, and this indeed independent of $\gamma>1$, because whatever one might get from a right-hand side $f$ is overruled by the decay one has for the Cauchy problem. I've edited the second part of my answer. | |
| Apr 20, 2015 at 6:27 | history | edited | ifw | CC BY-SA 3.0 |
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| Apr 19, 2015 at 12:08 | comment | added | Juhana Siljander | Ok, so this gives a condition on $p$ given $q$ and $d$. Is it known that this is optimal, i.e. for high enough $p$ there is, in general, no decay (independent of the decay rate $\gamma$ of the forcing term)? | |
| Apr 17, 2015 at 12:33 | comment | added | ifw | @Juhana $\sideset{_{2\,}}{_1}{F}(\dots)$ is the hypergeometric function. Otherwise, it is an exact evaluation of this integral (up to a multiplicative constant $C_{d,r}$). The condition on the function under the integral to be integrable is $\displaystyle \frac1q-\frac1p = 1-\frac1r<\frac2d$. | |
| Apr 17, 2015 at 11:05 | comment | added | Juhana Siljander | Thanks for answering my question. So what is this function $F$ and, in particular, how did you obtain the estimate where this function appears the first time? A simple integration would seem to give a restriction for $r$ in terms of $d$, in order the right hand side to be integrable. | |
| Apr 15, 2015 at 15:24 | history | edited | ifw | CC BY-SA 3.0 |
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| Apr 15, 2015 at 15:19 | history | answered | ifw | CC BY-SA 3.0 |